So despite what everybody here says, it is the property of the H- ion that gives the photosphere its opacity.

According to him it has nothing to do with distance in a thin plasma.

And I was right about the density of an object is what a blackbody spectrum is related to.

Is he right?

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I dont think that anyone here has ever truly discussed what causes opacity in a gas. This also misses the point on what the photosphere is in the first place.

I dont know how many times you need it explained to you but I will try again:

The photosphere is defined as the region where the gas making up the sun becomes optically thin, allowing radiation to finally escape the sun.

Like Tusenfem has explained a couple times, the blackbody radiation curve comes from the gas below the photosphere, not the photosphere itself.

If you think that he is saying that a gas cannot make a blackbody radiation curve, then he is wrong. Any optically thick object gives off a blackbody curve

First of all, just like I have said before and korjik, the black body spectrum is not produced by the photosphere, just as Freedman says. The photophere is not thick enough AND does not exist in LTE as most of the photons will escape from this region. But you just decide to misinterprete what we are writing on this board and take a face value (and misinterprete again, but I am not surprised) what someone else tells you.

Freedman is right and I am right and Korjik is right, however Upirver's interpretation of what these three people write is wrong.

Indeed, there is het H- ion, which creates opacity in the photosphere, but that has nothing to do with the BB spectrum of the sun. Opacitiy or optical thickness does have to do with how long the distance is if only because of the definition of optical thickness. Freedman does not say that the distance does not count, in this message you copied. He says There are enough H- ions in the photosphere to effectively absorb all colors of visible light, which is what makes the photosphere so opaque. This is meant in a column of the photosphere. You can imagine it yourself that if you have a 2 meter long column of thin gass, the change that a photon is absorbed is twice compared with a 1 meter long column, because it will have twice as much particles along the line of flight of the photon.

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All astronomers know quite well that the H- ion gives the photosphere its opacity. Freedman is not saying anything new here.
He does not say that, nor does he mean that. What he says is that the relevant distance depends on both the opacity per particle and the density of particles, so the presence of H- means that the distance you need can be achieved at a much lower density than you would otherwise need. The relevant distance for the photosphere to appear opaque is called the "scale height" of the photosphere, and is determined entirely by gas physics (it is the distance a proton at the temperature of the photosphere can travel against the Sun's gravity before falling back down). That the scale height is narrow is why the Sun's edge appears sharp. That we see this edge at a relatively low density is entirely due to the high opacity of H-, and this latter point is what Freedman is making (again, very well understood).

The other point he is making is that scattering opacity is bad for blackbody emission, you want the opacity to absorb the light and thermalize it, then re-radiate it based on the temperature. That way you get photons at the appropriate temperature, rather than "remembering" their old temperature. So high densities help that to happen in general, but H- opacity also does that pretty well, helping us to have a blackbody. That's the density issue, but there are other ways to do it at low density.

The connection between density and blackbodies is what I said-- the greater tendency to thermalize the light rather than simply scatter it. But there is also an indirect connection. A key question is, how much does the temperature change over the distance the light can penetrate? If that distance is small (i.e., high density or high opacity), then the temperature will not vary a lot and we have a good blackbody.

However, there is a more subtle point here that Freedman is not necessarily wrong about, but he doesn't seem to appreciate. In the case of the photosphere, the temperature structure is controlled by the radiation, not the other way around. But what is tricky is that the actual temperature of the gas will depend on how much the radiative opacity tends to scatter light, and how much it tends to thermalize light. So if there were no H-, the temperature structure in the layers we could see could still be the same, and so would the basic blackbody spectrum we see, as long as some other thermalizing opacity (albeit much smaller) would still dominate. The main thing that would be different if the opacity was much lower is that the photosphere would be found at a higher density, such that the depth light could penetrate would still be the same as it is now, and the temperature structure over that depth would also be the same (because the temperature is slave to the radiation). At those higher densities, the photosphere might still be pretty good at thermalizing the light, so I'm not convinced the spectrum would be much different at all without H-, in gross terms. The Sun would still be a 6000 K star even without H-, but there would be detailed changes in the spectrum relating to the higher surface density, most notably the broadness of spectral lines and the clarity of the ionization edges.

This is not "despite" what everybody has said here. It is in fact a repetition of what everybody has (or should have) said here.

Actually, according to him, yes it does. The more H^- there is along the optical path, the higher the opacity. A longer path means more H^-. Therefore, a longer path ("distance in a thin plasma") means greater opacity.

Well, since I don't know what you said or where you said it, maybe you were. The fact that anything that is dense (optically thick) enough yields a black body spectrum has never been contested.

On the matter of where the BB spectrum comes from, let me reinforce what others have already said.

... At the bottom of a stellar photosphere, the optical depth of the surface is high enough to prevent the escape of most photons. They are re-absorbed not far from where they are emitted. There are nearly as many photons headed into the star as there are headed outward. The material is then close to thermodynamic equilibrium, and we expect the radiation laws of a black body to describe the radiation in these deep layers of a stellar photosphere. Higher layers deviate increasingly from the black-body case as the leakage becomes more significant. There is a continuous transition from near perfect local thermodynamic equilibrium (LTE) deep in the photosphere to complete non-equilibrium (non-LTE) high in the atmosphere.
The Observation and Analysis of Stellar Photospheres, David F. Gray, Cambridge University Press, 2005 (3rd edition), ch. 6, p. 119.

The basic picture is fairly simple. The black body spectrum welling up from the deep sun is intercepted, and deformed, by the photosphere. The detailed picture is more complicated because LTE plasmas are much easier to deal with than are non-LTE plasmas. One should not interpret this with a flippant assumption that non-LTE plasmas are deeply mysterious, so "anything is possible". One should interpret this as meaning that those who are willing to do the work can deal with them effectively, and those who are not willing to do the work, cannot deal with them effectively.

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500 kilometers of air is moderately opaque as illustrated by rather dim sunlight typical just before sun set. At that thickness, trace impurities reduce the transparency. Neil

And, ironically, that is at much higher densities than the Sun's photosphere-- yet the opacity you are talking about is scattering opacity. Thermalizing opacity would make the daytime sky as dark as night! So again we see that the main importance of H- is to thermalize, not just to be opaque, even for very low densities.

I don't wish to take this thread off-topic, but I'm wondering if another of the great successes in modelling stars could be introduced here - the instability strip and the role of opacity in creating variability in the delta Scuti's, RR-Lyraes and Cepheids?

I think it's a great story, and illustrates well the role of specific atomic (and ionic) species in such obviously macroscopic phenomena, explanations that would be impossible without atoms whose behaviour is explicable only with quantum theory.

Wont you have to ban yourself if you continue to take the thread off topic and Upriver reports you?

It really is amazing just how well stars are modelled tho. The explanation of the instability strip, and its identification in stars from white dwarfs to supergiants is a good one. Type Ia supernova are another good one. Most thought that Ias were due to collapse at the chandrasekar limit till someone ran the numbers and found that it should be runaway carbon burning.

I had thought this allowed only a 200km deep photosphere, but apparently ~ 400km to 500km is the more popularly reported depth. Yet, if we use the temperature found at the bottom of the photosphere of 6390K and the top (limb) of 5000K, calculate the radiant energy production for both to solve for the inner Solar radius, the result is horribly off.

[E1*A1 = E2*A2; E = (Stefan-Boltzmann constant) * T^4, A = 4 pi r^2]

Using either the Solar Planck temp. of (5850K) or the Solar effective temp. (5777K) produces a Photosphere depth of over 100,000 km.

So what gross error am I making this time?

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The formula you use applies only at one place, the depth to which we see. It can be used if you know the radius r and surface temperature T to find the luminosity of the star. There's no sense to which you can write it at two different places and equate them-- it only gives the correct luminosity if applied at the single correct place (the visible surface). I'm not sure what you mean about the 200 km vs. 500 km-- the scale height is more like the former as I recall.

This is where I'm clearly confussed. The upper "surface" could be argued to be 5000K since it is the temperature we see at the limb, and the lower "surface" (bottom of photosphere) is 6390K. The assumption (which is where I am likley off-track) is the higher temperature found on the bottom will be as a result of the amount of energy coming from a smaller surface area but having a total energy equal to the upper photosphere surface energy; the total energy flux always being the same regardless of radius.

However, if my prior formulation and logic were correct (not!), then at a 500km depth the temp. difference would be only about 2K. Yet, there is some logic in the comparison of energy flux of the various inner surface shells with this quirky temp. result that might prove helpful in understanding more about opacity and scale heights and any other issue that deserves a little more understanding.

Some have stated a 200km depth for the photosphere, but, now that I look a little deeper, 400km (NASA) and 500km (Big Bear Obs.) depths seem to be the more common estimates.

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Those are not really two different surfaces-- if the Sun were really a blackbody, they would be the same surface. That they are not, and are at two different temperatures, is really just a measure of the degree to which the Sun is not an ideal blackbody. Ideal blackbodies don't have limb darkening, for example.

Yes, that assumption is off track. The temperature difference is significant, the radius difference is negligible. It's just a measure of the limitation of our idealization here-- the actual "surface" that applies for the "effective temperature" is somewhere in between.
I think this just depends on the fairly arbitrary definition of the photosphere. The most physically meaningful depth is the scale height, of about 200 km. But if one wants to specify some range, like the distance from the temperature minimum to the tau=2/3 point, you can get more-- but it's just a convention about our labels, the physics is in the scale height.

Yes, that is where my train seems to be running out of track. Yet, since the energy production is a given, it seems logical that for any radius outside the core, this flux divided by the surface area would produce its own Planck temperature; the closer to the core, the hotter the temp., and the greater the scale height.

Yes, but they come pretty close. No doubt, selecting any one small region would reveal a better bb curve, the Sun's sp. irr. is usually the light from the entire disk and, thus, an integrated curve.

Yes, that makes good sense as the effective temperature (5777K) is close to a bb in distribution a (5850K being the best fit Planck temp.), and it falls between the 5000K limb temp. and the hotter appearing central zone temperature of 6390K.

Yes, I get 193km using the 6390K and a surface gravity of 274m/s^2 (which may be a value after c.f.). At 5000K, it would be 151km, assuming I've gotten it right.

If I understand scale height, and I don't much, admittedly, it is the height a column of fluid must be for a given temperature such that it's kinetic energy would match the gravitational force upon it. How this ties to what we see optically is not clear to me.

That's got to be true.

The confussing issue, which I create (no charge, of course ), seems to be the idea that if we did a cut-a-way of the sun, we could calculate Planck temperatures at any radius point since we know the energy that must be moving through the imaginary spherical shell at that point.

Why it doesn't work this way is still unclear. Of course, it doesn't remotely work that way in our atmosphere even though air isn't too bad of an ideal gas. Our atmosphere actually gets hotter in places higher up, colder in regions down below (when looking from above). Yet, I doubt the Sun has any such odd thermal layers (though convection temp. gradients must exist).

[Added: One reason I like this equal power per imaginary shell is because it works so nicely once radiation leaves the surface. Take the flux per meter at the surface, then expand the radius to 1 a.u. (is it A.U. or a.u.?, it's not a proper name, but I see a lot of A.U.s) and... voila.... 1366 w/m^2. ]

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Keep in mind that this formula is arrived at by integrating the Planck Function over all frequencies (or all wavelengths), -infinity to +infinity. So it is true if and only if the radiated spectrum is a true blackbody. But that will not be true anywhere, except perhaps at the very bottom of the photosphere. Absorption & scattering in the photosphere (but especially absorption I think) will redistribute energy as a function of frequency (or wavelength), therefore seriously deforming the original Planck spectrum. So I am not surprised if the simple Stefan-Boltzmann equation does not measure up to expectations.

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The surface of last scattering.

From the Swedish solar observatory;

"The base of the triangles aligns with the dark "floor" of the small sunspot while the peak indicates the apparent height of the small "light-bridges" of granular features projecting into the sunspot. The measured height is between 200 and 450 km with an average of about 300 km and an uncertainty of 50 km."
http://www.lmsal.com/Press/SPD2003.html

How could those peaks be brighter than the valleys which are closer to the heat?
The peaks are as high as the estimate of the photospheres thickness...
http://www.lmsal.com/Press/Lites_tri..._disptri24.jpg

Below the photosphere.
So the whole spectrum comes from one specific level?
Or different levels fro different wavelengths?

I would love to see that laboratory experiment producing a blackbody with plasma.

Where I used to work they are work getting a plasma spectrum at many times higher pressures.
The Nature paper spectrum is taken at 2.5 atm drive pressures which produces 8000 atm in the bubble at stagnation.
The current drive pressure at the lab is running around 2500 atm. I cant wait till they get a spectrum.

I know for sure if you pressurize a plasma to 8000 atm, it gives off a quasi continuum that looks nothing like a BB, ask Korjik(Nature article).
What level is 8000 atm in the sun?

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Yes, but stars do come pretty close. [I am curious what bodies are known as better bb objects. I doubt any hole in any furnace I cut into would be much good. ]

Here is a comparison of the Sun's spectral irradiance from space with that of two Planck temperatures: the effective temperature (5777K) and the Planck temperature (~5850K). The former is the one used most often since it is the Planck temperature that would match the integrated power across all wavelengths. Yet, as we know, since the Sun is not a bb, a 5850K Planck temperature will more closely match the sp. irr. of the Sun for most of the wavelengths. That is not a lot of difference in Planck temperature.

Yes, and the outer atmosphere, especially the chromosphere, I think, acts more like a cloud than a radiator, creating absorption bands, though slight.

However, my 1.7K (2K) Stefan-Boltzmann approach is way off from the actual 1390K center to limb temperature variation observed. My train has not made the wrong turn, it has run off the track instead. So, my problem must not be in the minor bb variation.

Ken's answer in scale height has got to be it, but it hasn't cleared up my shell bias, yet. This will require even more georgeeze than normal, no doubt. Perhaps for the outer layer, the only real temperature rule for the gas is the balance between gravity and KE (ie scale height).

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Gee, I dont know, maybe it is because sunspots are cooler and so the bright spot is actually hotter?

One of the problems that you have is that you seem to think that pictures tell all. The surface that you are looking at is the surface of last scattering that you mentioned above. Now since the rest of us realize that the surface of the sun is gas, and actually watch movies that show the gas convecting, and moving, and since some of us also realize that since sunspots are cooler they sink, some of us realize that the surface we are seeing actually moves. Hot parts rise, cool parts sink and all that stuff.
technically, if you really really wanted to, you could find the exact opacity per unit wavelength, but I doubt it would help much. The variation in seeable depth wont be much compared to the thickness of the sun.

You've shown a graph with that at least twice now
technically, all that is irrelevant. The optical thickness is what is important, not the pressure. You still dont know the dirrerence between smeared emission lines and blackbody either

What a lot of people don't know is that the flux = sigma*T^4 formula only applies at the surface of the blackbody. In the interior, if there's no T gradient, there's no flux at all, or for the Sun, there has to be a flux, so there is a T gradient. The flux is the same, the T gets higher and higher-- the formula becomes meaningless below the surface. It is also meaningless above the surface in the chromosphere, because the local T goes up there but the flux stays the same. This is a surface formula, for one surface only, and if T varies in the region of interest, it's just not an ideal blackbody.
You have to be careful, because at 5000K the hydrogen goes neutral and you get half as many particles and half as much pressure. That makes the scale height half what it would be for ionized hydrogen. Another way to look at this is, the scale height is the height the protons go before falling back down, but if there are free electrons, which are very fast of course, gravity alone could not pull them back down-- there's also an electric field needed to keep the electons from flying off into space (their scale height would normally want to be 1800 times more). It must work out that the acceleration of that field on the electrons is half that of the gravity on the protons, so that the combined effect of gravity and electric field on the protons is also half the acceleration of gravity alone. That allows the protons and electrons to accelerate together, as needed for charge neutrality, and it effectively makes gravity half as strong on the protons, and the scale height twice as high. So I think ionized H gives you 400 km, and neutral gives more like 200 km. The photosphere we normally think of, where there's lots of H-, has to be fairly neutral.
It is true that the flux stays fixed, what does not stay true is that you can equate the flux to sigma*T^4. That formula assumes if you look in one direction, you see a blackbody at T, and if you look the other, you see empty space. Thus above the photosphere, the T is not the local T, it is the T of the photosphere, and below the photosphere, the assumption that you see empty space when you look up breaks down. Thus the formula only applies right at the surface, and when the appropriate surface is not so straightforward to specify, then the formula is only approximate.

If you are talking about continuum emission, then yes it's probably just hot gas being moved around. But there's another issue here, called the chromosphere, which was indeed a giant surprise when it was first discovered. The chromosphere is above the photosphere, and its "brightness" is not seen at all frequencies, only at very special frequencies where the opacity goes way up. Here we are not talking about H- opacity any more, which applies over most of the visible spectrum, but rather only very special frequencies where the low-density atoms undergo transitions between energy levels. These are called "spectral lines", and to see the brightening you are talking about, you look in those spectral lines. The reason you see a brightening when you do that is because the chromosphere is actually hotter than the photosphere, even though it is farther from the heat, just as you say. That was the mystery, and the resolution appears to be that the chromosphere is mechanically heating by shock waves and by magnetic fields, emanating up from the convection zone beneath the photosphere (where it is very hot).
It's pretty tough because low density plasmas are highly scattering. And even if you did get a blackbody spectrum to come out of it, your plasma would still not "be a blackbody", however, because the flux it emitted into free space outside the plasma would be well below sigma*T^4, due to all that scattering. Nevertheless, if you get enough of it, all at the same T, and go deep inside it, you will see a thermal radiation field at that T, but no net flux at all. That's an important difference between being inside and being outside an obect at T.
A crucial question here is not just what the density is, but whether or not its optically thick. It probably isn't. Nevertheless, it is indeed still true that even something that is optically thick and all at the same temperature does not have to emit a blackbody spectrum into free space outside the object, if there is scattering opacity. And if the scattering opacity is wavelength-dependent relative to the thermalizing opacity, you won't even get the shape of a blackbody curve.

Most of what you are saying is right, but it's not true that only the optical thickness matters. Pressure matters too, in that it tends to control the ratio of scattering to thermalizing opacity. Scattering opacity will generally interfere with the production of an ideal blackbody spectrum, and tends to be more important at low pressure.

Yes, I see now. The idea of applying an inverse square law radiatve transfer to a convective gas is very erroneous. If it were to apply, using the outer core "surface" temp. of 7,100,000K, the Planck temperature at the current visible surface would be 3,800,000K (ignoring all the changes this would cause the Sun to go through). [Global Warming would be re-defined quickly. ]

Could I use this approach for the radiative zone?

The convective zone acts like an insulator to the flux; it keeps the inner regions nice and toasty. Other math tools and models are used to handle it, scale height being a good one, no doubt.

Interesting. Sounds like it expands to twice its volume when this recombination occurs, correct? I assume this greatly affects the opacity of the outer photosphere.

Yes, that makes good sense since scale height is kinetic energy / gravitational retention. How this might tie to opacity is not clear to me, though.

A 3d model of electron traffic should be quite interesting. Since scale height is inverse to particle mass, 1800x difference makes good sense, also.

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It isn't obvious to me that the scale height of the Sun's
photosphere defines the photosphere's depth. The top, yes,
if it defines the limit of how far up protons are tossed by
thermal agitation before they begin to fall back under gravity,
but not the bottom.

As far as I can tell, the photosphere is defined by the region from
which thermal radiation escapes into space. The top is where the
highest particles emit light, and the bottom is where the lowest
particles emit light that is able to escape without being absorbed
by another particle. No light at all reaches us from below the
bottom of the photosphere, by definition.

The bottom of the photosphere is no different from the region
above it or the region below it, except that the density varies
smoothly with radius.

There are a bunch of different reasons why the Sun's radiation
curve is not a perfect blackbody curve. One of them is the fact
that light reaches us from throughout the thickness of the
photosphere, where the temperature varies with radius. Fitting
a single curve to the combined light is a useful approximation,
but of no fundamental significance. If a single layer of the
photosphere could be viewed, then a better blackbody curve
would be seen, though it would of course still suffer from other
effects such as thermal Doppler shifts, line absorption, and
fluorescence.

-- Jeff, in Minneapolis

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I was limiting myself to what he was mentioning. In the case where you are in a lab experiment, the optical thickness determines what you detect.

No. The problem is not the convection, the problem is the isotropic character of the radiation field. In such situations, the net flux does not depend on T, it depends on the gradient of T. Indeed, the need to carry a flux is just why the surface is so much cooler than the core. The fact that the surface has a larger surface area is a fairly minor detail, the main reason it cools is not the increasing surface area, it is the requirement that there be a temperature gradient to carry an internal flux (especially in the radiative zone, where this is all that is happening).

Radiative zones would keep the inner regions nice an toasty too. Indeed, convection is often a worse insulator than radiation-- that's the whole reason we have a convective zone. The convective zone is precisely the region where convection carries heat better than radiation does. The reason it carries heat better in that region is because that is the region that is unstable to convection, so a little convection will cause a lot of convection to appear, and before long it is carrying energy like gangbusters and takes over that role.
Yes.
Basically, the connection is that light in order to escape, light must be able to travel through the scale height. If it can't go that far, it won't escape, and will not be seen by us on Earth. If it can travel farther than that, it will be seen, but it is coming from less dense layers where there is not much light being emitted in the first place. The place that combines lots of light emission with a high fraction of that light getting out is the place where the distance light can travel (its mean free path) equals a scale height. That's what we call "the surface" of the Sun.

Yup, that's just where it comes from-- two particles with the same kinetic energy can go "up" different distances, in inverse proportion to their mass (as the energy is converted into mgH).

Yes, I didn't really explain that part, but see my comment to George just above. What one means by "the photosphere's depth" is actually just a convention, I'm talking about the width of the layer where most of the light we see comes from, which is also the region where H- opacity is so important. Some people might define the photosphere as the entire layer where energy is carried radiatively and not convectively, so that would extend quite a bit deeper into layers from which virtually no light emerges.

That is the definition I'm using, yes-- the region from which a significant majority of the thermal radiation emerges, i.e., the layer we "see".
Right. But note that this will be a scale height, because deeper than that, the light doesn't get out very well.
There's never any obvious place to draw such a line as "no light at all", we will always have to settle for most of the light fails to get out.

No, the degree of scattering will be quite important for what you detect. One must always distinguish the optical thickness for scattering from that for thermalization.

What do you mean by scattering opacity and thermalizing opacity?

Ok. Could I liken the radiative zone to a conductor with a given thermal conductivity constant?

So the randomly walking photons not only migrate but in the convective zone they get to ride, with their hot hosts, the vertical express train, too.

Well, I won’t say you’ve just scratched the surface , but there does seem to be more to all this. This almost sounds like the light we see comes from emissions from atoms that have risen up to this layer we call the surface. If so, then the scale height would determine the deepest layer they could have come from. Yet, this process as a light source must not be true else there would be no center to limb temp. variation. It must be some H ion species connection to how light is handled that limits the visible depth.

I would be curious of the flux variation with depth. Since the lower layers are less dense, as you say, it makes sense that the two Planck temperatures we use for the Sun are closer to the temperature at the bottom of the photosphere than the 5000K temp. at the top.

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Scattering opacity is opacity that does little more than change the direction that the photons are propagating, or from a wave perspective, it just sends out waves at the same frequency and phase that they come in. The local temperature is irrelevant. Thermalizing opacity first destroys the photon, then later re-emits radiation to conserve energy, but based on the local temperature. So for example, free electrons primarily just scatter, but H- opacity thermalizes, because the extra electron is knocked clear of the H atom, and is later replaced by some other electron from the thermal pool.